Natural and step response RLC find iL(t)

Click For Summary
The discussion focuses on solving for the inductor current iL(t) in an RLC circuit after a switch is closed at t=0. The circuit is simplified to a parallel configuration with a 4A current source, 2kΩ resistor, 62.5H inductor, and 2.5μF capacitor. Initial conditions are established with iL(0-) at -5mA and iL(∞) at 4mA, leading to the calculation of the natural frequency and damping factor. The characteristic equation yields roots that inform the solution for iL(t), expressed as iL=4-12e^-40t+3e^-160t for t≥0. The thread also notes a discrepancy regarding the circuit diagram and the problem statement.
pokie_panda
Messages
35
Reaction score
0

Homework Statement




The switch in the circuit in the figure has been open a long time before closing at t=0.
Find iL(t) t>0
Express your answer in terms of t, where t is in milliseconds.

https://www.flickr.com/photos/84781786@N03/8899251427/

The Attempt at a Solution



We simplify to circuit to get

4A current , 2kohms,62.5H,2.5uF (All in parallel with each other).

iL(0-)= -15/3000 =-5mA ,
Vc(0-)=0,
iL(∞)= 4mA,

wo^2= 1/LC=10^6/62.5*.5 =6400
wo=80 rad/s

a= 1/2RC =10^6/4000*2.5 =100

s1,s2=-100±√100^2-80^2=-100±60
s1=-40 , s2=-160

iL=If+A'1e^-40t + A'2e^-160t

iL(∞)=If=4mA
iL(0)=A'1+A'2+If=-5mA

so A'1+A'2=-9mA

diL/dt(0) =0=-40A1 - 160A'2

A'1=12mA A'2=3mA

iL=4-12e^-40t+3e^-160t t≥0
 

Attachments

  • Screen Shot 2013-05-29 at 9.45.23 PM.png
    Screen Shot 2013-05-29 at 9.45.23 PM.png
    4.5 KB · Views: 652
Last edited:
Physics news on Phys.org
Your diagram doesn't appear to match your problem statement.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K